Let's analyze and solve the problem given:

We have two functions:

• $f(x) = \sqrt{1 - x^2}$
• $g(x) = \sin(x)$ for $0 \leq x \leq \pi$

The questions are:

Part (a)

Determine the range of $g(x)$ and the domain of $f(x)$.

Solution:

1. Range of $g(x)$: Since $g(x) = \sin(x)$ and $x$ is in the interval $[0, \pi]$, the range of $\sin(x)$ over this interval is $[0, 1]$.

2. Domain of $f(x)$: For $f(x) = \sqrt{1 - x^2}$ to be defined, the expression inside the square root, $1 - x^2$, must be non-negative. Thus, $-1 \leq x \leq 1$. Therefore, the domain of $f(x)$ is $[-1, 1]$.

Part (b)

Show that $f \circ g$ is defined and determine its expression.

Solution:

• The composition $f \circ g$ means we substitute $g(x)$ into $f(x)$.
• So, $(f \circ g)(x) = f(g(x)) = f(\sin(x)) = \sqrt{1 - (\sin(x))^2}$.

Using the Pythagorean identity, we know that $1 - \sin^2(x) = \cos^2(x)$, so: $(f \circ g)(x) = \sqrt{\cos^2(x)} = |\cos(x)|$ Since $0 \leq x \leq \pi$, $\cos(x)$ is non-negative for $0 \leq x \leq \frac{\pi}{2}$ and non-positive for $\frac{\pi}{2} \leq x \leq \pi$. Therefore: $(f \circ g)(x) = \cos(x) \text{ for } 0 \leq x \leq \frac{\pi}{2}, \text{ and } -\cos(x) \text{ for } \frac{\pi}{2} \leq x \leq \pi.$

Part (c)

Determine the domain and range of $f \circ g$.

Solution:

1. Domain of $f \circ g$: Since $g(x)$ is defined for $0 \leq x \leq \pi$ and the range of $g(x)$ is $[0, 1]$, which falls within the domain of $f(x)$, the domain of $f \circ g$ is $0 \leq x \leq \pi$.

2. Range of $f \circ g$: The range of $f(g(x)) = |\cos(x)|$ over $0 \leq x \leq \pi$ is $[0, 1]$ because $\cos(x)$ varies between $1$ and $-1$ in this interval, and taking the absolute value gives values from $0$ to $1$.

Summary

• (a) Range of $g(x)$: $[0, 1]$; Domain of $f(x)$: $[-1, 1]$.
• (b) $f \circ g(x) = |\cos(x)|$.
• (c) Domain of $f \circ g$: $[0, \pi]$; Range of $f \circ g$: $[0, 1]$.

Would you like further explanations on any of these steps?

Here are five related questions that could extend your understanding:

1. What would happen to the composition $f \circ g$ if the range of $g(x)$ exceeded the domain of $f(x)$?
2. How would the answer change if $g(x) = \sin(x)$ were defined over a different interval, like $[0, 2\pi]$?
3. Can we determine $g \circ f$? If so, what would its domain and range be?
4. How does the Pythagorean identity help in simplifying compositions of trigonometric functions?
5. What is the significance of absolute values in the composition $f \circ g$?

Tip: When dealing with compositions, always ensure that the range of the inner function fits within the domain of the outer function.